Exercise 3.2.11 (If $H \leq K \leq G$, then $|G:H| = |G : K|\cdot |K : H|$)

Proof. Let ${(x_i)}_{i\in I}$ a complete family of left coset representatives of $K$ in $G$, and ${(y_j)}_{j\in J}$ a complete family of left coset representatives of $H$ in $K$, so that

Then

Indeed, if $g\in G$, there is some $i\in I$ and some $k\in K$ such that $g=x_{i}k$, and there is some $j\in J$ and some $h\in H$ such that $k=y_{j}h$, thus $g=x_i{y}_{j}h$ ,where $h\in H$, thus $g\in x_i{y}_{j}H$ for some $(i,j)\in I\times J$, so $g\in \underset{(i,j)\in I\times J}{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}{x}_i{y}_{j}H$. This show that $G\subseteq \underset{(i,j)\in I\times J}{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}{x}_i{y}_{j}H$, and since $x_i{y}_{j}H\subseteq G$ for all $(i,j)\in I\times J$, the converse inclusion is true. This show the equality (2).

Moreover, if $(i,j)\in I\times J,(k,l)\in I\times J$, then

To prove (3), suppose that $g\in x_i{y}_{j}H$ and $g\in x_k{y}_{l}H$.

Then $g\in x_{i}K$ and $g\in x_{k}K$, so $x_{i}K\cap x_{k}K\ne \varnothing$. Therefore $i=k$ by the first disjoint union in (1). Then $x_i^{-1}g\in y_{j}H$ and $x_i^{-1}g\in y_{l}H$, thus $y_{j}H\cap y_{l}H\ne \varnothing$. Therefore $j=l$ by (1), so $(i,j)=(k,l)$. This proves (3).

Hence ${(x_i{y}_j)}_{(i,j)\in I\times J}$ is a complete family of coset representatives of $H\cap K$ in $G$.

If $I$ or $J$ in infinite, then $I\times J$ is infinite, and $|G:H|=|G:K|\cdot |K:H|=\infty$.

If $I$ and $J$ are finite sets, $I\times J$ is finite and $\mathrm{Card}(I\times J)=\mathrm{Card}(I)\cdot \mathrm{Card}(J)$, where $\mathrm{Card}(I)=|G:K|,\mathrm{Card}(J)=|K:H|$ and $\mathrm{Card}(I\times J)=|G:H\cap K|$, therefore

□